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Integers from 1 to 25 (both inclusive) are split into 5 groups of 5 numbers each. The medians of these 5 groups are \(\mathrm{A}\), \(\mathrm{B}\), \(\mathrm{C}\), \(\mathrm{D}\) and \(\mathrm{E}\). If the average of these medians is \(\mathrm{m}\), what is the smallest values \(\mathrm{m}\) can take?
Let's break down what we're asked to do step by step. We have the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25 - that's all 25 numbers from 1 to 25.
We need to split these 25 numbers into exactly 5 groups, with each group containing exactly 5 numbers. So we'll have:
For each group, we find the median (the middle value when the 5 numbers are arranged in order). Let's call these medians A, B, C, D, and E.
Finally, we calculate the average of these 5 medians: \(\mathrm{m} = (\mathrm{A} + \mathrm{B} + \mathrm{C} + \mathrm{D} + \mathrm{E}) ÷ 5\)
Our goal is to find the smallest possible value for m.
Process Skill: TRANSLATE - Converting the word problem into clear mathematical requirements
This is the key insight that will drive our strategy. When we have any 5 numbers arranged in ascending order, the median is always the middle number - the 3rd number in the sequence.
For example, if a group contains the numbers \(\{2, 7, 11, 18, 23\}\), then arranged in order they are: 2, 7, 11, 18, 23, and the median is 11.
Here's what this means for our strategy: if we want to control what the median of a group will be, we need to ensure that exactly 2 numbers in that group are smaller than our chosen median, and exactly 2 numbers are larger.
So if we want the number 'x' to be the median of a group, we need to:
To minimize the average \(\mathrm{m} = (\mathrm{A} + \mathrm{B} + \mathrm{C} + \mathrm{D} + \mathrm{E}) ÷ 5\), we want to make the sum A + B + C + D + E as small as possible.
This means we want each individual median (A, B, C, D, E) to be as small as possible.
The smallest numbers we have available are 1, 2, 3, 4, and 5. Let's see if we can make these five numbers our five medians.
Can 1 be a median? For 1 to be a median, we'd need 2 numbers smaller than 1 in its group. But there are no numbers smaller than 1 in our list. So 1 cannot be a median.
Can 2 be a median? For 2 to be a median, we'd need 2 numbers smaller than 2. We only have the number 1. So 2 cannot be a median.
Can 3 be a median? For 3 to be a median, we'd need 2 numbers smaller than 3. We have 1 and 2. So 3 CAN be a median! The group would be \(\{1, 2, 3, ?, ?\}\) where the question marks are numbers larger than 3.
Following this logic:
So our medians should be 3, 4, 5, 6, 7 to achieve the minimum sum.
Let's construct the actual groups to verify our strategy works:
Group 1: \(\{1, 2, 3, 24, 25\} → \mathrm{median} = 3\)
Group 2: \(\{4, 5, 6, 22, 23\} → \mathrm{median} = 6\)
Wait, let me reconsider this more systematically. I want to minimize the sum of medians, so let me try to make the medians 3, 4, 5, 6, 7:
Group 1: \(\{1, 2, 3, 4, 5\} → \mathrm{median} = 3\)
But this uses up numbers 1, 2, 4, and 5, which I needed for other medians. Let me think more strategically.
If my medians are A, B, C, D, E and I want to minimize their sum, I should try:
Group 1: \(\{1, 2, 3, 23, 25\} → \mathrm{median} = 3\)
Group 2: \(\{4, 5, 6, 22, 24\} → \mathrm{median} = 6\)
Group 3: \(\{7, 8, 9, 20, 21\} → \mathrm{median} = 9\)
Group 4: \(\{10, 11, 12, 18, 19\} → \mathrm{median} = 12\)
Group 5: \(\{13, 14, 15, 16, 17\} → \mathrm{median} = 15\)
Sum of medians = \(3 + 6 + 9 + 12 + 15 = 45\)
Average = \(45 ÷ 5 = 9\)
Let me verify this uses all numbers from 1 to 25: ✓
Process Skill: APPLY CONSTRAINTS - Ensuring we use each number exactly once while achieving our optimization goal
The smallest possible value for m is 9.
This matches answer choice C.
To verify: We arranged the numbers so that the medians are 3, 6, 9, 12, and 15, giving us an average of \((3+6+9+12+15)÷5 = 45÷5 = 9\). Any attempt to make the medians smaller would require having medians of 1 or 2, which is impossible since they don't have enough smaller numbers to be in the middle position of a 5-number group.
1. Misunderstanding what constitutes a median in a group of 5
Students often confuse the median with the average or think the median is just any middle number. They may not realize that in a group of 5 numbers, the median is specifically the 3rd number when arranged in ascending order, requiring exactly 2 numbers below it and 2 numbers above it.
2. Not recognizing the constraint that limits which numbers can be medians
Students may attempt to use very small numbers like 1 or 2 as medians without realizing that there aren't enough smaller numbers available to support them as medians. For example, trying to make 1 a median when there are no numbers smaller than 1 to fill the required positions.
3. Focusing on minimizing individual medians rather than their sum
Students might get fixated on trying to make each median as small as possible individually, rather than understanding that they need to minimize the total sum of all five medians while respecting the constraints of using each number from 1 to 25 exactly once.
1. Incorrect grouping that violates the 'each number used exactly once' constraint
When constructing the five groups, students may accidentally reuse numbers or forget to include certain numbers, leading to invalid arrangements. For instance, they might put the number 4 in multiple groups or completely omit number 17 from all groups.
2. Arithmetic errors when calculating the median positions or final average
Students may miscalculate which position represents the median in an ordered group of 5, or make computational errors when adding the medians (3+6+9+12+15) or dividing by 5 to get the final average.
3. Creating groups where the intended median is not actually in the middle position
Students might construct a group intending for a specific number to be the median, but fail to properly arrange the other numbers in the group, resulting in a different number being the actual median.
1. Selecting a value that represents the sum of medians rather than their average
After correctly finding that the sum of optimal medians is 45, students might select 45 as their answer instead of dividing by 5 to get the average of 9. This is especially likely if they lose track of what the question is actually asking for.